【发布时间】:2018-03-19 12:02:20
【问题描述】:
我正在查询 3 个表,并且可以使用单个查询对每个表提取正确的数据,当我将它们放入连接语句时,返回的结果是不同的。
查询:
select advisor, round(sum(shift),1) AS avail_time from shift_track
where date >= 20180319
group by advisor order by advisor ASC
select advisor, round(sum(time/60),1) AS standard_time from shift_report
where date >= 20180319
group by advisor order by advisor ASC
select advisor, round(sum(time/60),1) AS mobile_time from sr_mobile
where date >= 20180319
group by advisor order by advisor ASC
加入我正在尝试:
SELECT shift_track.advisor, shift_track.date, sum(shift_track.shift) AS
avail_time,
sum(shift_report.time/60) AS stand_time,
sum(sr_mobile.time/60) AS mobile_time
from shift_track
inner join shift_report on shift_report.advisor=shift_track.advisor
inner join sr_mobile on sr_mobile.advisor=shift_track.advisor
where shift_track.date =20180319
group BY shift_track.advisor ORDER BY shift_track.advisor ASC
我正在尝试从每个表格中获取顾问在给定日期的总时间。加入查询似乎跟随堆栈上的其他人。
另一种方法是在 PHP 中使用单独的查询,然后将结果添加到表中以显示 - 但我确信我应该能够在单个查询中完成它??
非常感谢任何帮助。
【问题讨论】:
标签: mysql inner-join